Đáp án:

$\begin{array}{l}
a)5.\sqrt {{{\left( { - 2} \right)}^4}}  = 5.\sqrt {16}  = 5.4 = 20\\
\sqrt {11 - 6\sqrt 2 }  + \sqrt {6 - 4\sqrt 2 } \\
 = \sqrt {9 - 2.3.\sqrt 2  + 2}  + \sqrt {4 - 2.2.\sqrt 2  + 2} \\
 = \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}}  + \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
 = 3 - \sqrt 2  + 2 - \sqrt 2 \\
 = 5 - 2\sqrt 2 \\
b)2\sqrt {{{\left( { - 2} \right)}^4}}  + 3.\sqrt {{{\left( { - 2} \right)}^8}} \\
 = 2.4 + 3.16\\
 = 56\\
\sqrt {6 + 2\sqrt {4 - 2\sqrt 3 } }  = \sqrt {6 + 2\sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} } \\
 = \sqrt {6 + 2\left( {\sqrt 3  - 1} \right)} \\
 = \sqrt {4 + 2\sqrt 3 }  = \sqrt 3  + 1\\
c)\sqrt {{{\left( {4 - 3\sqrt 2 } \right)}^2}}  + \sqrt {17 + 12\sqrt 2 } \\
 = 4 - 3\sqrt 2  + \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} \\
 = 4 - 3\sqrt 2  + 3 + 2\sqrt 2 \\
 = 7 - \sqrt 2 \\
\sqrt {24 + 8\sqrt 5 }  + \sqrt {9 - 4\sqrt 5 } \\
 = \sqrt {{{\left( {2\sqrt 5  + 2} \right)}^2}}  + \sqrt {{{\left( {\sqrt 5  - 2} \right)}^2}} \\
 = 2\sqrt 5  + 2 + \sqrt 5  - 2\\
 = 3\sqrt 5 \\
d)\sqrt {12 - 6\sqrt 3 }  = \sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}}  = 3 - \sqrt 3 \\
\sqrt {6 - 4\sqrt 2 }  + \sqrt {22 - 12\sqrt 2 } \\
 = \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}}  + \sqrt {{{\left( {3\sqrt 2  + 2} \right)}^2}} \\
 = 2 - \sqrt 2  + 3\sqrt 2  + 2\\
 = 2\sqrt 2  + 4\\
e)\sqrt {11 - 6\sqrt 2 }  - \sqrt {6 - 4\sqrt 2 } \\
 = \left( {3 - \sqrt 2 } \right) - \left( {2 - \sqrt 2 } \right)\\
 = 1\\
\sqrt {17 - 12\sqrt 2 }  + \sqrt {9 + 4\sqrt 2 } \\
 = \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}}  + \sqrt {{{\left( {2\sqrt 2  + 1} \right)}^2}} \\
 = 3 - 2\sqrt 2  + 2\sqrt 2  + 1\\
 = 4\\
B1)\\
1)DKxd: - 3x + 7 \ge 0\\
 \Leftrightarrow 3x \le 7\\
 \Leftrightarrow x \le \dfrac{7}{3}\\
Vậy\,x \le \dfrac{7}{3}\\
2)Dkxd:9 - 12x + 4{x^2} > 0\\
 \Leftrightarrow {\left( {3 - 2x} \right)^2} > 0\\
 \Leftrightarrow x \ne \dfrac{3}{2}\\
Vậy\,x \ne \dfrac{3}{2}\\
3)Dkxd:\dfrac{{x + 2}}{{4 - x}} \ge 0\\
 \Leftrightarrow \dfrac{{x + 2}}{{x - 4}} \le 0\\
 \Leftrightarrow  - 2 \le x < 4\\
Vậy\, - 2 \le x < 4\\
4)\dfrac{{x + 1}}{{5x + 2}} \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x > \dfrac{{ - 2}}{5}\\
x \le  - 1
\end{array} \right.\\
Vậy\,x \le  - 1\,hoac\,x >  - \dfrac{2}{5}\\
5)Dkxd:\left\{ \begin{array}{l}
{x^2} - 4 \ge 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ {x \ge 2} \right.\\
Vậy\,x \ge 2\\
6)\left\{ \begin{array}{l}
5 - x \ge 0\\
x + 1 > 0\\
9 - {x^2} \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 - 1 < x \le 5\\
{x^2} \le 9
\end{array} \right.\\
 \Leftrightarrow  - 1 < x \le 3\\
Vậy\, - 1 < x \le 3\\
7)\left( {3 - x} \right)\left( {x + 2} \right) \ge 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) \le 0\\
 \Leftrightarrow  - 2 \le x \le 3\\
Vậy\, - 2 \le x \le 3\\
8){x^2} - 6x \ge 0\\
 \Leftrightarrow x\left( {x - 6} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 6\\
x \le 0
\end{array} \right.\\
Vậy\,x \le 0\,hoac\,x \ge 6
\end{array}$